Cho hàm số \(f\left( x \right)\)...

B

- Hướng dẫn giải

Phương pháp giải:

Tích phân 2 vế \(f\left( x \right)f'\left( x \right) + 18{x^2} = \left( {3{x^2} + x} \right)f'\left( x \right) + \left( {6x + 1} \right)f\left( x \right)\,\,\forall x \in \mathbb{R}\).

Giải chi tiết:

Ta có :

\(f\left( x \right).f'\left( x \right) + 18{x^2} = \left( {3{x^2} + x} \right)f'\left( x \right) + \left( {6x + 1} \right)f\left( x \right)\)\(\begin{array}{l} \Rightarrow \int\limits_0^x {\left( {f\left( x \right).f'\left( x \right) + 18{x^2}} \right)} dx = \int\limits_0^x {\left( {\left( {3{x^2} + x} \right)f'\left( x \right) + \left( {6x + 1} \right)f\left( x \right)} \right)} dx\\ \Leftrightarrow \int\limits_0^x {f\left( x \right).f'\left( x \right)} dx + \int\limits_0^x {18{x^2}} dx = \int\limits_0^x {{{\left( {\left( {3{x^2} + x} \right)f\left( x \right)} \right)}^\prime }} dx\\ \Leftrightarrow \left. {\left( {\dfrac{1}{2}{{\left( {f\left( x \right)} \right)}^2} + 6{x^3}} \right)} \right|_0^x = \left. {\left( {\left( {3{x^2} + x} \right)f\left( x \right)} \right)} \right|_0^x\\ \Leftrightarrow \left( {\dfrac{1}{2}{{\left( {f\left( x \right)} \right)}^2} + 6{x^3}} \right) - \left( {\dfrac{1}{2}{{\left( {f\left( 0 \right)} \right)}^2} + 0} \right) = \left( {3{x^2} + x} \right)f\left( x \right) - 0\\ \Leftrightarrow {\left( {f\left( x \right)} \right)^2} - 2\left( {3{x^2} + x} \right)f\left( x \right) + 12{x^3} = 0\\ \Leftrightarrow {\left( {f\left( x \right)} \right)^2} - 2\left( {3{x^2} + x} \right)f\left( x \right) + 12{x^3} = 0\\ \Leftrightarrow {\left( {f\left( x \right)} \right)^2} - 2\left( {3{x^2} + x} \right)f\left( x \right) + {\left( {3{x^2} + x} \right)^2} = {\left( {3{x^2} + x} \right)^2} - 12{x^3}\\ \Leftrightarrow {\left( {f\left( x \right) - \left( {3{x^2} + x} \right)} \right)^2} = {\left( {3{x^2} - x} \right)^2}\\ \Leftrightarrow \left[ \begin{array}{l}f\left( x \right) - \left( {3{x^2} + x} \right) = 3{x^2} - x\\f\left( x \right) - \left( {3{x^2} + x} \right) =  - 3{x^2} + x\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}f\left( x \right) = 6{x^2}\\f\left( x \right) = 2x\end{array} \right.\end{array}\)

Do \(f\left( 0 \right) = 0,\,\,f'\left( 0 \right) \ne 0\) nên \(f\left( x \right) = 2x\). Khi đó:

 \(\begin{array}{l}\int\limits_0^1 {\left( {x + 1} \right){e^{f\left( x \right)}}dx = } \int\limits_0^1 {\left( {x + 1} \right){e^{2x}}dx = \dfrac{1}{2}} \int\limits_0^1 {\left( {x + 1} \right)d\left( {{e^{2x}}} \right)} \\ = \left. {\dfrac{1}{2}\left( {x + 1} \right).{e^{2x}}} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {{e^{2x}}d\left( {x + 1} \right)}  = \left. {\dfrac{1}{2}\left( {x + 1} \right).{e^{2x}}} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {{e^{2x}}dx} \\ = \left. {\dfrac{1}{2}\left( {x + 1} \right).{e^{2x}}} \right|_0^1 - \left. {\dfrac{1}{4}{e^{2x}}} \right|_0^1 = \dfrac{1}{2}.2.{e^2} - \dfrac{1}{2}.1.{e^0} - \dfrac{1}{4}{e^2} + \dfrac{1}{4}{e^0} = \dfrac{7}{4}{e^2} - \dfrac{1}{4}\\ \Rightarrow a = \dfrac{7}{4};\,\,b =  - \dfrac{1}{4}\,\,\, \Rightarrow a - b = 2\end{array}\)

Chọn: B